Question

id An Olympic archer misses the bul's-eye 14 % of the time. Assume each shot is independent of the others. If she shoots 7 arrows, what is the probability of each of the results described in parts a through f below? The probability is 0.0766 (Round to four decimal places as needed.) O b) She misses the bul's-eye at least once. The probability is 0.6521 nl (Round to four decimal places as needed.) c) Her first miss comes on the second or third arrow The probability is 0.2239 (Round to four decimal places as needed.) d) She misses the bull's-eye exactly 3 times. The probability is V (Round to four decimal places as needed.)

Answer 1

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solletion! Given n=7, P=0.14 q-l-p-1-0.14 -0.86 -a) P(first miss in 7th) = 0.86 x 0.14 20.05663941 20.0566 b) PEEins o miss at least once) at P (no miss )= 1-0.867 21-0.34792782222 = 0.652072177 0.6521 c) P(first miss in and our 3rd) = P(first miss in and)t P (first miss in ard) 0.86*0:14 +0-86*0.14 -0.1200+ 0.103544 = 0.293944 20.9239 d) P(x=3) binomial distribution P(x-3)có 88 o vu ° F (0 0% 253444 P(x-3) > 0 059 5.

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